| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Expected frequency with unknown parameter |
| Difficulty | Standard +0.3 This is a standard two-part normal distribution problem requiring inverse normal calculation to find σ, then a forward probability calculation with expected frequency. While it involves multiple steps and parameter estimation, the techniques are routine for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z = 1.015\) | B1 | Accept \(z\) between \(\pm1.01\) and \(1.02\) |
| \(1.015 = \dfrac{70 - 69}{\sigma}\) | M1 | Standardising |
| \(\sigma = 0.985\ (200/203)\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(58 + 9 = 67\) | M1 | \(58 + 9\) seen or implied (or \(69-58\) or \(69-9\)) |
| \(P(>67) = P\!\left(z > \dfrac{67-69}{0.9852}\right)\) | M1 | Standardising \(\pm z\); no cc; allow their sd (must be \(+\)ve) |
| \(= P(z > -2.03) = 0.9788\) | M1 | Correct prob area |
| \(300 \times 0.9788\) | M1 | Multiply their prob (from use of tables) by 300 |
| \(= 293.6\) so 293 | A1 [5] | Accept 293 or 294 from fully correct working |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 1.015$ | **B1** | Accept $z$ between $\pm1.01$ and $1.02$ |
| $1.015 = \dfrac{70 - 69}{\sigma}$ | **M1** | Standardising |
| $\sigma = 0.985\ (200/203)$ | **A1** [3] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $58 + 9 = 67$ | **M1** | $58 + 9$ seen or implied (or $69-58$ or $69-9$) |
| $P(>67) = P\!\left(z > \dfrac{67-69}{0.9852}\right)$ | **M1** | Standardising $\pm z$; no cc; allow their sd (must be $+$ve) |
| $= P(z > -2.03) = 0.9788$ | **M1** | Correct prob area |
| $300 \times 0.9788$ | **M1** | Multiply their prob (from use of tables) by 300 |
| $= 293.6$ so **293** | **A1** [5] | Accept 293 or 294 from fully correct working |
---5 The heights of school desks have a normal distribution with mean 69 cm and standard deviation $\sigma \mathrm { cm }$. It is known that 15.5\% of these desks have a height greater than 70 cm .\\
(i) Find the value of $\sigma$.
When Jodu sits at a desk, his knees are at a height of 58 cm above the floor. A desk is comfortable for Jodu if his knees are at least 9 cm below the top of the desk. Jodu's school has 300 desks.\\
(ii) Calculate an estimate of the number of these desks that are comfortable for Jodu.
\hfill \mbox{\textit{CAIE S1 2016 Q5 [8]}}